Let $R$ be a regular local ring, and let $P$ be a prime ideal in $R$. Assume that $\operatorname{height}(P)=h$ and $P=(x_1,\ldots, x_k)$, for some $x_1,\ldots,x_k\in R$, and $k\geq h$. Let $J=(x_1,\ldots, x_h)$ be an ideal in $R$ of height $h$, namely $\operatorname{height}(J)=h=\operatorname{height}(P)$. I would like to show that $P$ is minimal over $J$.
I know that by the converse implication of Krull's Height Theorem, any prime ideal of height $h$ is minimal over an ideal generated by $h$ elements. However, it does not look like one can choose arbitrarily these $h$ elements to be the first $h$ elements out of the generating set of $P$.
I would be happy enough if one can choose any $h$ elements out of the generating set of $P$, say $y_1,\ldots, y_h\subseteq \{x_1,\ldots, x_k\}$ such that $P$ is minimal over $(y_1,\ldots,y_h)$.
Thank you for any help!
I suppose that $R$ is a Cohen-Macaulay local ring. Then the height of $J$ equals its grade. Since $J$ is generated by $h$ elements, and grade of $J$ is $h$, $x_1,\dots,x_h$ is a regular sequence. It follows that $x_1,\dots,x_h$ is a maximal regular sequence in $P$, and therefore $P$ is associated to $R/(x_1,...,x_h)$. But $R/(x_1,...,x_h)$ is also Cohen-Macaulay, so $P$ is minimal over $(x_1,...,x_h)$.